Find values for the real constants and
so that the following limit equation holds:
Incomplete.
Find values for the real constants and
so that the following limit equation holds:
Incomplete.
Find values for real numbers and
so that the following improper integral equation holds:
First, we have
In order for this to converge we must have the coefficient of in the numerator equal to 0; hence, we must have
. Now, to solve the integral equation we have
But, by assumption this integral equals 1, so we have
Since we then have
Find a value for the constant so that the improper integral
converges and find the value of the integral in this case.
First, we have
For this to converge we must have the coefficient of the in the numerator going to 0 as
(otherwise the integral will converge by limit comparison with
). Hence,
Now, we need to evaluate the integral. Incomplete.
Find the value of so that the improper integral
converges, and compute the value of the integral in this case.
First, we compute the value of the constant ,
This integral will diverge by comparison with if the coefficient of the
term in the numerator is not equal to zero. Hence, we must have
Next, we evaluate the integral with ,
Find a value for the constant so that the improper integral
converges. Find the value of the integral for this value of .
First, we compute the value of ,
Since this integral will diverge by limit comparison with for any nonzero coefficient of the
term in the numerator we must have
Next, we evaluate the integral
Test the following improper integral for convergence:
The integral converges if and diverges for
.
Proof. We can compute this directly. We evaluate the indefinite integral (using the substitution ) for
:
But then the limit
is finite for and diverges for
(since
as
and so the limit diverges when
and converges for
). Hence, the integral
converges for and diverges for
. In the case that
the indefinite integral is
and as
, so the integral divers for
as well. Therefore, we have the integral converges for
and diverges for
Test the following improper integral for convergence:
The integral diverges.
Proof. First, we show that for all
. To do this let
This derivative is 0 at and is less than 0 for
and greater than 0 for
. Hence,
has a minimum at
. But,
(since
implies
). So,
has a minimum at
and is positive there; thus, it is positive for all
, or
So, since we know
But then, consider the limit
Therefore, by the limit comparison test (Theorem 10.25), the convergence of would imply the convergence of
(for any
), but we know by Example 5 that this integral diverges. Hence, we must also have the divergence of
Test the following improper integral for convergence:
The integral converges.
Proof. First, we write
For the first integral we know for all we have
; hence,
Then, the integral
(We know by L’Hopital’s, writing
or by Example 2 on page 302 of Apostol.) Hence,
converges by the comparison theorem (Theorem 10.24 on page 418 of Apostol).
For the second integral, we use the expansion of about
,
Then we have
But this integral converges since it has no singularities.
Thus, we have established the convergence of
Test the following improper integral for convergence:
The integral converges.
Proof. We can compute the value of the improper integral directly. First, we can evaluate the indefinite integral using integration by parts with
Therefore,
So, to evaluate the improper integral we take the limit,
Test the following improper integral for convergence:
The integral converges.
Proof. We can compute this integral directly. First, we evaluate the indefinite integral using the substitution ,
.
Now, we have discontinuities at both limits of integration so we evaluate by taking two limits,